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COMMUNICATION THEORY
7CCEMCTH
COURSEWORK
There are 3 Questions, answer all.
Detailed answers and calculations are required.
Upload clearly scanned copies of your written answers by
the deadline, as indicated on Keats.
1
Question 1 - Infinite-length design for QAM channel equalisation [40 marks]:
The discrete-time deterministic autocorrelation function of a filtered AWGN channel is given
as:
() = 0.5−1 + 1 − 0.5
The signal-to-noise ratio matched filter bound is SNRMFB = 50 and
ℰ̅
2
= 25, both in linear
scale, where ℰ̅ is the average transmit energy per dimension and
2 is the noise variance.
Successive QAM symbols with average transmit power =
ℰ
= 0dBm, where is the
symbol period, are transmitted through this channel.
The noise variance is 2 = −80dBm/Hz. The gap at symbol error probability of = 10
−6 is
Γ = 8.8dB.
a) Design MMSE-LE for this channel by finding its filter transform and compute the
corresponding unbiased signal-to-noise ratio. (15 marks)
b) Design MMSE-DFE for this channel by finding its filter transforms and compute the
corresponding unbiased signal-to-noise ratio. (15 marks)
c) Determine which one of your designs in parts (a) and (b) can provide the highest
data rate with integer number of bits per symbol at the target probability of symbol
error of = 10
−6, and compute that maximum achievable data rate on this
channel. (10 marks)
2
Question 2 – Finite-length design [20 marks]:
Consider a filtered AWGN channel with impulse response modelled as:
ℎ() = () − .5( − ).
The transmit filter (basis function) is given as () = √
1
sinc (
), where is the symbol
period. The noise variance is 2 = 0.125, the average transmit energy per dimension is
ℰ̅ = 1. An oversampling factor of = 1 is assumed. The impulse response of the anti-
aliasing filter at the receiver is given as ℎ() =
1
sinc (
), i.e., flat in the frequency
domain with a gain of one.
Hint: sinc (
+
) ∗ sinc (
+
) = sinc (
+(+)
), where ∗ means convolution.
Important attention: The objective of this question is to explore the calculations of the
formulations in lecture notes in details. You may use MATLAB for any matrix manipulations
required for completing your solution. You may also validate your detailed solutions and
answers with the MATLAB DFE programme in KEATS, after you have solved the problem and
written down your solutions in details.
a) Find the pulse response () of the channel for finite-length design and the power
gain ‖‖2, corresponding to the discrete-time channel. (6 marks)
b) Find the signal-to-noise ratio matched filter bound, SNRMFB (in dB). (2 marks)
c) Design a 3 tap FIR MMSE-LE for delay Δ = 0. (8 marks)
d) Find the −
2 for your design in part (c). (2 marks)
e) Compute the unbiased SNR for the MMSE-LE. (2 marks)
3
Question 3 – Finite-length design and evaluations with oversampling and polyphase
channel modelling [40 marks]:
Consider a filtered AWGN channel with impulse response ℎ() =
1
1
1+(
107
3
)2
and the
frequency response (Fourier transform of ℎ()):
() =
3
10
−6(10
−7)||.
QAM transmission with symbol rate of 1 MHz and carrier frequency = 600 KHz is used on
this channel with oversampling factor of 2. The transmission system is shown in Figure 1:
Figure 1. Detailed transmission channel
As shown on Fig. 1, the power spectral density of noise is −86.5 dBm/Hz and the transmit
power is 1mW. The oversampling factor for the design of the equaliser is = 2. Square root
raised cosine pulse () with 10% of excess bandwidth (i.e., roll-off factor = 0.1) is used
as the transmit basis function. An ideal anti-aliasing filter with frequency response as
characterised on Fig. 1 is applied at the receiver input.
A MATLAB program provided on KEATS (Project 1) can be used to implement ().
According to the lecture notes the complex pulse response is modelled as the cascade of the
transmit basis function (), baseband equivalent of the channel impulse response,
ℎ(), and the impulse response of the ideal anti-aliasing filter, ℎ():
() = () ∗ ℎ() ∗ ℎ().
a) Using MATLAB programming, find the complex discrete-time pulse response samples
and write down as your answer for part (a) only the 8 samples around the peak value
(i.e., centred around time origin = 0 , with time zero included), as the truncated
pulse response for practical design.
(20 marks)
Hint: Using MATLAB, first find the continuous-time Fourier transform of () in
baseband for which you will need to find ( + ) =
3
10
−6(10
−7)|+|. You will
also need square root raised cosine pulse in frequency domain, i.e., √() in lecture
notes, whose MATLAB program can be found in KEATS (Project 1). With = 2, your
sampling period will be
2
. Then you will be able to find the discrete-time Fourier
transform (DTFT) of the oversampled discrete-time pulse response. Then using
inverse DTFT, i.e., -point IFFT in MATLAB, you will be able to find the samples of
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the pulse response. To avoid aliasing in discrete-time domain, use large enough,
i.e., = 1024. After obtaining discrete-time samples of pulse response, truncate
it around the peak of the pulse response, i.e., around the time origin, and keep 8
samples.
Here is a minimum programming steps, suggested for part (a):
T = 1*10^(-6); % Symbol Period
fc = 6*10^(5); % Carrier Frequency
alpha = 0.1; % Excess Bandwidth
l=2; %oversampling factor
N = 1024; % Number of FFT points
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Form raise cosine pulse in frequency-domain, i.e. √(),
as the continuous-time Fourier transform of ().
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Form the continuous-time channel baseband spectrum, i.e.,
( + ) =
3
10
−6(10
−7)|+|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Form the continuous-time equivalent channel pulse
response for finite-Length in frequency-domain, i.e., ()
as the continuous-time Fourier transform of
() = () ∗ ℎ() ∗ ℎ().
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Form the discrete-time Fourier-transform (DTFT) of the
discrete-time sequence = (
2
).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Use the N-point IFFT command to find the discrete-time
samples of the pulse response .
Use the ifftshift command to centre time samples about
the time origin
Record 8 time samples of (4 samples to the left of the
time origin, sample at the origin and 3 samples to the
right of the time origin).
Continue Question 3 with the following remaining parts, which are independent from Part
(a):
• The discrete-time complex pulse response at carrier frequency = 550 KHz, the symbol
rate of 1 MHz for QAM transmission through the transmission system shown and detailed in
Fig. 1 with oversampling factor of = 2 is given as:
() = (0.022 − 0.053)−4 + (0.097 + 0.03)−3 + (−0.046 + 0.1)−2 +
(−0.23 − 0.15)−1 + 0.4 + (−0.23 + 0.15) + (−0.046 − 0.1)2
+ (0.97 − 0.03)3
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b) After applying delay to make causal, express the obtained samples as causal 2-
tuple representation of the polyphase channel pulse response, i.e., calculate the
matrix = [0 1 ⋯ ]. (5 marks)
c) Calculate channel power gain ‖‖2 in dB. (3 marks)
Hint: Use ‖‖2 =
1
2
∑ |(
2
)|
2
d) Calculate signal-to-noise ratio matched filter bound. (2 marks)
e) Using the DFE programme in KEATS, design an FIR MMSE-LE with 10 taps and find
the corresponding achievable data rate at probability of symbol error of 10−6.
(5 marks)
f) Using the DFE programme in KEATS, design an FIR MMSE-DFE with 10 feedforward
taps and 4 feedback taps and find the corresponding achievable data rate at
probability of symbol error of 10−6.